tìm x biết:
a,13/x-1+5/2x-2-6/3x-3
b,2x-3/3+-2/3=5-3x/6-1/3
các bn giúp mk nha,bn nào lm xong trước chiều nay mk sẽ cảm ơn các bn rất nhìu
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Những câu hỏi liên quan
31 tháng 3 2018
Bài 1 :
Ta có :
\(\left|2x-1\right|=5\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=6\\2x=-4\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{6}{2}\\x=\frac{-4}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)
Vậy \(x=-2\) hoặc \(x=3\)
Bài 2 :
Đặt \(A=\frac{3x+4}{x-1}\) ta có :
\(A=\frac{3x+4}{x-1}=\frac{3x-3+7}{x-1}=\frac{3x-3}{x-1}+\frac{7}{x-1}=\frac{3\left(x-1\right)}{x-1}+\frac{7}{x-1}=3+\frac{7}{x-1}\)
Để A là số nguyên thì \(\frac{7}{x-1}\) phải nguyên \(\Rightarrow\)\(7⋮\left(x-1\right)\)\(\Rightarrow\)\(\left(x-1\right)\inƯ\left(7\right)\)
Mà \(Ư\left(7\right)=\left\{1;-1;7;-7\right\}\)
Suy ra :
| \(x-1\) | \(1\) | \(-1\) | \(7\) | \(-7\) |
| \(x\) | \(2\) | \(0\) | \(8\) | \(-6\) |
Vậy \(x\in\left\{-6;0;2;8\right\}\) thì \(A\inℤ\)
Chúc bạn học tốt ~
3 tháng 7 2018
a. Ta có: \(x^2-10x+26+y^2+2y=0\Leftrightarrow\left(x^2-10x+25\right)+\left(y^2+2y+1\right)=0\\ \)
\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2=0\Rightarrow\hept{\begin{cases}x+5=0\\y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}}\)
b. \(\left(2x+5\right)^2-\left(x-7\right)^2=0\Leftrightarrow\left(2x+5+x-7\right).\left(2x+5-x+7\right)=0\)
\(\Leftrightarrow\left(3x-2\right).\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x+12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-12\end{cases}}}\)
c. \(25.\left(x-3\right)^2=49.\left(1-2x\right)^2\Leftrightarrow\left(5x-15\right)^2=\left(7-14x\right)^2\Leftrightarrow\left(5x-15\right)^2-\left(7-14x\right)^2=0\)
\(\Leftrightarrow\left(5x-15-7+14x\right).\left(5x-15+7-14x\right)=0\Leftrightarrow\left(19x-22\right).\left(-9x-8\right)=0\)
\(\Leftrightarrow\left(19x-22\right).\left(9x+8\right)=0\Leftrightarrow\orbr{\begin{cases}19x-22=0\\9x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{22}{19}\\x=-\frac{8}{9}\end{cases}}}\)
d. \(\left(x+2\right)^2=\left(3x-5\right)^2\Leftrightarrow\left(x+2\right)^2-\left(3x-5\right)^2=0\Leftrightarrow\left(x+2+3x-5\right).\left(x+3-3x+5\right)=0\)
\(\Leftrightarrow\left(4x-3\right).\left(8-2x\right)=0\Leftrightarrow\orbr{\begin{cases}4x-3=0\\8-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=4\end{cases}}}\)
e. \(x^2-2x+1=16\Leftrightarrow\left(x-1\right)^2-16=0\Leftrightarrow\left(x-1-4\right).\left(x-1+4\right)=0\)
\(\Leftrightarrow\left(x-5\right).\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
16 tháng 3 2020
1) Ta có: \(4x^2-1=\left(2x+1\right).\left(3x-5\right)\)
\(\Leftrightarrow\left(2x+1\right).\left(2x-1\right)-\left(2x+1\right).\left(3x-5\right)=0\)
\(\Leftrightarrow\left(2x+1\right).\left[\left(2x-1\right)-\left(3x-5\right)\right]=0\)
\(\Leftrightarrow\left(2x+1\right).\left(2x-1-3x+5\right)=0\)
\(\Leftrightarrow\left(2x+1\right).\left(4-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\-x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\left(TM\right)\\x=4\left(TM\right)\end{matrix}\right.\)
Vậy \(x=-\frac{1}{2}\) hoặc \(x=4\)
2) Ta có: \(\left(x+1\right)^2=4.\left(x^2-2x+1\right)\)
\(\Leftrightarrow\left(x+1\right)^2-\left[2.\left(x-1\right)\right]^2=0\)
\(\Leftrightarrow\left[\left(x+1\right)+2.\left(x-1\right)\right].\left[\left(x+1\right)-2.\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+1+2x-2\right).\left(x+1-2x+2\right)=0\)
\(\Leftrightarrow\left(3x-1\right).\left(3-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\-x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(TM\right)\\x=3\left(TM\right)\end{matrix}\right.\)
Vậy \(x=\frac{1}{3}\) hoặc \(x=3\)
3) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x.\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x.\left(2x^2-x+6x-3\right)=0\)
\(\Leftrightarrow x.\left[x.\left(2x-1\right)+3.\left(2x-1\right)\right]=0\)
\(\Leftrightarrow x.\left(x+3\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-3\left(TM\right)\\x=-\frac{1}{2}\left(TM\right)\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=-3\) hoặc \(x=-\frac{1}{2}\)
4) Ta có: \(2x=3x-2\)
\(\Leftrightarrow2x-3x=-2\)
\(\Leftrightarrow-x=-2\)
\(\Leftrightarrow x=2\left(TM\right)\)
Vậy \(x=2\)
5) Ta có: \(x+15=3x-1\)
\(\Leftrightarrow x-3x=-1-15\)
\(\Leftrightarrow-2x=-16\)
\(\Leftrightarrow x=8\left(TM\right)\)
Vậy \(x=8\)
6) Ta có: \(2-x=0,5x-4\)
\(\Leftrightarrow-x-0,5x=-4-2\)
\(\Leftrightarrow-1,5x=-6\)
\(\Leftrightarrow x=4\left(TM\right)\)
Vậy \(x=4\)
16 tháng 3 2020
1) 4x2-1=(2x+1)(3x-5)
<=> (2x-1)(2x+1)-(2x+1)(3x-5)=0
<=> (2x+1)(2x-1-3x+5)=0
<=> (2x+1)(4-x)=0
<=>\([^{2x+1=0}_{4-x=0}< =>[^{2x=-1}_{x=4}< =>[^{x=\frac{-1}{2}}_{x=4}\)
2) (x+1)2= 4(x2-2x+1)
<=> x2+2x+1-4(x2-2x+1)=0
<=> x2+2x+1-4x2+8x-4=0
<=> -3x2+10x-3=0
<=> -3x2+x+9x-3=0
<=> -x(3x-1)+3(3x-1)=0
<=> (3x-1)(3-x)=0
<=> \([^{3x-1=0}_{3-x=0}< =>[^{3x=1}_{x=3}< =>[^{x=\frac{1}{3}}_{x=3}\)
3) 2x3+5x2-3x=0
<=> 2x(x2+\(\frac{5}{2}x-\frac{3}{2})=0\)
<=> 2x\(\left[x^2+2.\frac{5}{4}x+\frac{25}{16}-\left(\frac{25}{16}+\frac{3}{2}\right)\right]=0\)
<=> 2x\(\left[\left(x+\frac{5}{4}\right)^2-\frac{49}{16}\right]=0\)
<=> 2x\(\left(x+\frac{5}{4}-\frac{7}{4}\right)\left(x+\frac{5}{4}+\frac{7}{4}\right)=0\)
<=> x\(\left(x-\frac{1}{2}\right)\left(x+3\right)=0\)
<=>\(\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\\x=-3\end{matrix}\right.\)
4) 2x=3x-2
<=> 2x-3x=-2
<=> -x=-2
<=> x=2
5) x+15=3x-1
<=> x-3x=1-15
<=> -2x=-14
<=> x=-14:-2
<=> x=7
6) 2-x=0,5x-4
<=> -x-0,5x=-4-2
<=> -1,5x=-6
<=> x= -6: -1,5
<=> x=4
học tốt nghen
23 tháng 1 2018
câu a giống Võ Đoan Nhi
câu b:
( x2 + 2x -11 ) : ( x + 2)
=> x2 + 2x -11 : ( x + 2)
=> x(x+2) -11 : ( x + 2)
Vì x( x + 2) : ( x + 2) nên -11 : ( x + 2)
=> x + 2 thuộc ước của -11
ta lập bảng..............
VD
19 tháng 1 2018
\(3x+4⋮x-3\)
\(\Leftrightarrow3\left(x-3\right)+10\)\(⋮x-3\)
-Mà: \(3\left(x-3\right)⋮x-3\Rightarrow10⋮x-3\)
\(\Leftrightarrow x-3\inƯ\left(10\right)\Leftrightarrow x-3\in\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
-Lập bảng:.....
